Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
1
/
2 2
/ /
3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1
/
2 2
3 3
错误的解法，就是遇到多个 null 值的时候就会出问题，也就是在有的时候虽然他不对称，
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import org.junit.jupiter.api.Test;
import java.util.ArrayList;
import java.util.List;
public class
public boolean isSymmetric (TreeNode root)
List<TreeNode> list = new ArrayList<>();
inOrderTraversal(root, list);
int center = list.size() / 2 ;
return compare(list, center);
}
public boolean compare (List<TreeNode> list, int center)
for (int i = 0 , j = list.size() - 1 ; i < center; i++, j--) {
if (list.get(i).val != list.get(j).val) {
return false ;
}
}
return true ;
}
public void inOrderTraversal (TreeNode root, List<TreeNode> list)
if (root == null ) {
return ;
}
inOrderTraversal(root.left, list);
list.add(root);
inOrderTraversal(root.right, list);
}
void test ()
}
}
好的解法就是使用递归，所谓对称简单来说就是 root 的左孩子和右孩子要一样才行
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